Undamped and Damped Oscillations in Free Vibration

Last edited: 2024-11-08 22:13:16

Free vibration occurs when a spring-mounted system oscillates without external forces. Let us solve the formulas and equations of free vibration.

Undamped Free Vibration

Free vibration is the term for when a spring-mounted body is disturbed from its position of equilibrium. Undamped free vibration is ideal because all free vibration of particles, to a varying degree, dampens due to mechanical or fluid friction.

Undamped Free Vibration Formula

Looking at the figure above and doing a free-body diagram we can deduct using Newton's second law that

$$m \ddot{x} + kx = 0,$$

where $m$ is the body's mass, $x$ is the distance from equilibrium (when the force from the spring is 0), $\ddot{x}$ is the second time-derivative of $x$ and $k$ is the spring constant.

Undamped Free Vibration Equation Solved

Looking at the ODE in the equation above we can expect an oscillatory motion so we can write $x$ as

$$x = A \cos{(\omega_n t)} + B \sin{(\omega_n t)}.$$

Inserting this in the ODE we get:

$$m (-A \omega^2 \cos{(\omega_n t)} - B \omega^2 \sin{(\omega_n t)})$$ $$+ k (A \cos{(\omega_n t)} + B \sin{(\omega_n t)}) = 0.$$

By solving for the sine and cosine terms we get:

$$-m B \omega_n^2 + k B = 0 \text{ and } -m A \omega_n^2 + k A = 0.$$

Natural Circular Frequency Formula

So we only have a solution if

$$\omega_n = \sqrt{\frac{k}{m}},$$

which is called the natural circular frequency. Knowing that the expression for $x$ is the solution we can calculate $A$ and $B$. If we evaluate $x$ and $\dot{x}$ at $t=0$ we get $x_0 = A$ and $\dot{x} = B \omega$. Substituting these values into the solution we get

$$x = x_0 \cos{(\omega_n t)} + \frac{\dot{x}_0}{\omega} \sin{(\omega_n t)}.$$

Equilibrium Position For Vertical Spring

If we were to flip the system mentioned above we would also have the gravitational force so now the equilibrium position will contain a nonzero spring deflection $\delta_\text{eq}$. So now the equation for Newton's second law is

$$m \ddot{x} + k(\delta_\text{eq} + x) - mg = 0.$$

For the equilibrium position to be equilibrium the sum of all forces must be zero so

$$k \delta_\text{eq} = mg.$$

So the equation for Newton's second law becomes the formula that we had for the horizontal case:

$$m \ddot{x} + kx = 0.$$

So by defining the displacement variable to originate from the equilibrium, we can ignore the equal and opposite forces associated with equilibrium.

Damped Free Vibration

Let us introduce damping into the system in the form of a viscous damper that behaves linearly. The viscous damper consists of a cylinder filled with fluid and a piston with holes so the fluid can flow between the sides of the piston. The damping force's magnitude is proportional to the velocity of the body. The direction of the force is in the opposite direction of the velocity so we can describe the damping force on the body as $-c \dot{x}$. $c$ is called the viscous damping coefficient.

Damped Free Vibration Formula

Doing a free-body diagram on the body gives us this equation derived from Newton's second law:

$$m \ddot{x} + c \dot{x} + kx = 0.$$

Viscous Damping Factor / Damping Ratio

We discovered that for undamped free vibration $\omega_n = \sqrt{\frac{k}{m}}$. Let us make that substitution to the equation above and let us also make this substitution:

$$\zeta = \frac{c}{2 m \omega_n},$$

which is called the viscous damping factor or damping ratio which is a measurement of the severity of the damping. So now the equation becomes:

$$\ddot{x} + 2 \zeta \omega_n \dot{x} + \omega_n^2 x = 0.$$

Damped Free Vibration Equation Solved

To solve the equation above we assume a solution in the form of

$$x = A e^{\gamma t}$$

and substituting this into the equation gets us

$$\gamma^2 + 2 \zeta \omega_n \gamma + \omega^2 = 0.$$

This is called the characteristic equation and the roots for it is

$$\gamma_1 = \omega_n (-\zeta + \sqrt{\zeta^2 - 1})$$ $$\gamma_2 = \omega_n (-\zeta - \sqrt{\zeta^2 - 1})$$

Using the property of superposition due to the system's linearity we get the general solution which is the sum of the two individual solutions gained from the roots of the characteristic equation:

$$x = A_1 e^{\gamma_1 t} + A_2 e^{\gamma_2 t}$$ $$= A_1 e^{\omega_n (-\zeta + \sqrt{\zeta^2 - 1}) t} + A_2 e^{\omega_n (-\zeta - \sqrt{\zeta^2 - 1}) t}.$$

Categories of Damped Motion

The damping factor $\zeta$ can have values between 0 and infinity so the expression $(\zeta^2 - 1)$ can be positive, negative, or zero. This leads to three categories of damped motion.

Overdamped Motion

Overdamped motion is when $\zeta > 1$. This leads to the roots $\gamma_1$ and $\gamma_2$ being distinct from each other and them being real and negative. As a consequence, the equation of the solution above will approach zero when $t$ goes to infinity while not having any oscillations.

Critically Damped Motion

When $\zeta = 1$ the motion is critically damped and the roots are equal to each other:

$$\gamma_1 = \gamma_2 = - \omega_n.$$

The roots are also real and negative since $\omega$ is always positive. It can be shown that for critically damped motion the solution is given by

$$x = (A_1 + A_2 t) e^{-\omega_n t}.$$

A critically damped system will also decay with no oscillations to zero when $t$ goes to infinity and it will do so faster than an overdamped system. The figure below shows the difference between the two cases. They have the same initial conditions and $\zeta = 3$ for the overdamped system.

Underdamped Motion

The system is underdamped when $\zeta < 1$ so now $\zeta^2-1$ is negative and the solution becomes

$$x = A_1 e^{\omega_n (-\zeta + i \sqrt{1 - \zeta^2}) t} + A_2 e^{\omega_n (-\zeta - i \sqrt{1 - \zeta^2}) t}$$ $$= (A_1 e^{i \omega_n \sqrt{1 - \zeta^2} t} + A_2 e^{i \omega_n \sqrt{1 - \zeta^2} t}) e^{-\zeta \omega_n t},$$

since $\sqrt{-1} = i$. Let $\omega_d = \omega_n \sqrt{1 - \zeta^2}$ so

$$x = (A_1 e^{i \omega_d t} + A_2 e^{-i \omega_d t}) e^{-\zeta \omega_n t}.$$

By using the formula $e^{\pm i x} = \cos{x} \pm i \sin{x}$ we can rewrite the previous equation as

$$[(A_1+A_2) \cos{(\omega_d t)} + i (A_1 - A_2) \sin{(\omega_d t)}] e^{-\zeta \omega_n t}.$$

We define two new constants as $A_3 = (A_1 + A_2)$ and $A_3 = i (A_1 - A_2)$ so

$$x = [A_3 \cos{(\omega_d t)} + A_4 \sin{(\omega_d t)}] e^{-\zeta \omega_n t}.$$

Since $\cos{(\omega_d t)}$ and $\sin{(\omega_d t)}$ has the same frequency the expression above can be written as

$$x = Ce^{\zeta \omega_n t} \sin{(\omega_d t + \varphi)}.$$

To determine the constants $C$ and $\varphi$ you plug in the initial conditions for the system (for example $x(t=0) = x_0$ and $\dot{x}(t=0) = \dot{x}_0$).

Damped Natural Frequency and Damped Period

Analyzing our final formula above we can see it has an angular frequency of

$$\omega_d = \omega_n \sqrt{1 - \zeta^2}.$$

This angular frequency is called the damped natural frequency. Using the damped natural frequency we get the damped period:

$$\tau_d = \frac{2 \pi}{\omega_d}.$$

You can see an example in the figure below of an underdamped motion with $\zeta = 0.1$. The factor $C e^{-\zeta \omega_n t}$ dictates the amplitude of the motion.

Determining The Damping Ratio With Logarithmic Decrement

We can calculate the damping ratio if we were to measure two successive amplitudes (at for example $x_1$ and $x_2$) a damped period apart:

$$\frac{x_1}{x_2} = \frac{C e^{-\zeta \omega_n t_1}}{C e^{-\zeta \omega_n (t_1 + \tau_d)}} = e^{\zeta \omega_n \tau_d}.$$

Taking the natural logarithm of the previous equation we get the logarithmic decrement $\delta$:

$$\delta = \ln{\frac{x_1}{x_2}} = \zeta \omega_n \tau_d = \zeta \omega_n \frac{2 \pi}{\omega_n \sqrt{1 - \zeta^2}} = \frac{2 \pi \zeta}{\sqrt{1 - \zeta^2}}.$$

Now we can solve for the damping ratio:

$$\zeta = \frac{\delta}{\sqrt{4 \pi^2 + \delta^2}}.$$